Chapter 4.31 - Structure of Sulfur Hexafluoride

• We are discussing the structure of those molecules in which hybridization in the central atom involves d-orbitals also. In the previous section 4.30, we saw the structure of PCl₅. In this section, we will see SF₆

Structure of SF₆
• When we discussed VSEPR theory, we saw the Lewis dot structure of SF₆
    ♦ See fig.4.98(c) in section 4.16
    ♦ All the five S-F bonds are single bonds
• For convenience, that fig. is shown again below:

• The details about the shape of SF₆ can be written in 11 steps:
1. The S atom in SCl₆ is sp³d² hybridized
• Let us see how this sp³d² hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.181(a) below shows the orbitals of S (Sulfur)

Fig.4.181

(ii) When enough energy is given,
    ♦ one electron in the 3s orbital jumps to the 3d_i orbital
    ♦ one electron in the 3p_x orbital jumps to the 3d_ii orbital
• Thus we get six half filled orbitals. This is shown in fig.b
(iii) The 3s orbital mixes together with 3p_x, 3p_y, 3p_z, 3d_i and 3d_ii. This is also shown in fig.b
(iv) Since there is one s-orbital, three p-orbitals, and two d orbitals, it is a sp³d² hybridization
2. We know that:
    ♦ In sp³d² hybridization, there will be six resulting hybrid orbitals
• Together, they form a square bipyramidal (octahedral) shape
    ♦ This can be explained in 3 steps:
(i) Fig.4.182(a) below, shows the outlines of a octahedron

Fig.4.182

    ♦ The base square is shown in blue dashed lines
          ✰ The small red sphere is the nucleus of the S atom
          ✰ This nucleus is situated a the 'center of gravity' of the base square
    ♦ Four green dashed lines radiate out from the nucleus
          ✰ These lines are at an angular distance of 90^o apart
    ♦ Two red dashed lines radiate upwards and downwards from the nucleus
          ✰ These lines together form the axis of the octahedron
    ♦ The magenta dashed lines are the sloping edges of the octahedron
(ii) Lines which are required:
    ♦ We require the following lines:
          ✰ The four green dashed lines
          ✰ The two red dashed lines (these two, together form the axis)
    ♦ We do not require the following lines:
          ✰ The eight magenta dashed lines
          ✰ The four blue dashed lines
(iii) The required lines are shown in fig.4.182(b)
    ♦ Altogether, there are six lines
    ♦ The six hybrid orbitals will be oriented along these lines
    ♦ This is shown in fig.c
■ Thus we get the 'sp³d² hybridized S atom'
3. Let us write three 'important points to remember' about fig.4.182:
(i) The 'plane of the base square' is exactly at the middle of the structure
    ♦ So this plane is called 'equatorial plane'
(ii) All the four green dashed lines lie on the 'plane of the base square'
    ♦ So the four orbitals along the green lines are called 'equatorial orbitals' 
    ♦ These orbitals are at an angular distance of 90^o apart
(iii) The two red dashed lines form the axis of the structure
    ♦ So the two orbitals along the red lines are called 'axial orbitals' 
    ♦ They are at an angular distance of 180^o apart
4. So we can write:
    ♦ The 'sp³d² hybridized S atom' consists of two items:
          ✰ The six hybrid orbitals
          ✰ The nucleus (small red sphere)
 • In the fig.4.178(c), the 'smaller lobes of hybrid orbitals' are not shown
    ♦ This is to reduce congestion and thus obtain greater clarity
5. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp³d² hybrid orbitals are formed from 'one s orbital', 'three p orbitals' and 'two d orbitals'
(ii) In our present case of S atom, they are: 'one 3s orbital', 'three 3p orbitals' and 'two 3d orbitals'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of six electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the six electrons?
Answer:
• The six electrons will be distributed among the six sp³d² hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.182(c) above

6. So in fig.4.178(c), we have:

    ♦ Six half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. It can be written in 2 steps:
(We have already seen the two steps in the previous section and in the previous case of PCl₅. But we will write them again)
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron
• We saw it in fig.4.166 in the previous section. It is shown again below:

Fig.4.166

• The p_x orbital is shown in red color
    ♦ It has two arrows
• The p_y orbital is shown in green color
    ♦ It has two arrows
• The p_z orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
7. One Cl atom will come and overlap with each of the six half filled orbitals of the S atom
• So a total of six Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.183(a) below:

Fig.4.183

• In the fig.4.183(a) above, the 'p_z orbitals of Cl' have become completely filled
8. The electron clouds of Cl:
(i) The red, green, blue and cyan regions around the Cl nucleus are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(ii) So we do not need to show the orbitals (electron clouds) of the Cl atoms. We can delete them
• However, we will retain the p_z orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the S atom
(iii) The final structure is shown in fig.4.183(b) above
• The small yellow spheres represent the nuclei of Cl atoms

9. Thus we get the model of the SF₆ molecule
• Let us write the salient features of this model. It can be written in 3 steps:
(i) There is a total of six S-F bonds
    ♦ Four of them lie on the equatorial plane
          ✰ Along the diagonals of the base square
    ♦ The remaining two lie along the 'axis of the octahedron'
(ii) So there are:
    ♦ Four equatorial bonds
    ♦ Two axial bonds
(iii) The angle between
    ♦ Any two equatorial bonds is 90^o
    ♦ The two axial bonds is 180^o
    ♦ Any equatorial bond and axial bond is 90^o
10. The 2D representation is shown below:

• Recall that, we saw the same 2D representation when we analysed SF₆ using VSEPR theory. See fig.4.98(c) in section 4.16
• There we saw the reason for giving the solid triangle, dashed triangle etc.,
11. Note that, all the bonds in SF₆ are sigma bonds

---

• We have seen the a number of examples in which the central atom undergoes some sort of hybridization

• We have seen examples for sp³, sp², sp, sp³d and sp³d² hybridization
• Let us now write the salient features of hybridization
• We have already seen the application of those features in the various examples
• However, it is better to prepare a compilation of the features
• There are four salient features
1. The 'number of hybrid orbitals' is equal to the 'number of atomic orbitals that get hybridized'
Application example:
• In sp³d hybridization, five atomic orbitals are involved:
    ♦ One s orbital
    ♦ Three p orbitals
    ♦ One d orbital
• So the total number of participating orbitals is five 
• After the hybridization, we get five hybrid orbitals
2. The hybrid orbitals are always equivalent in energy and shape
Application example:
• Regarding shape:
    ♦ In all the examples, we see the same hybrid orbitals
          ✰ Each of those orbitals has a larger lobe and a smaller lobe
• Regarding energy:
    ♦ In CH₄, four hybrid orbitals are produced
          ✰ All four have the same energy
    ♦ In PCl₅, five hybrid orbitals are produced
          ✰ All five have the same energy
3. The hybrid orbitals are more effective in forming stable bonds than pure atomic orbitals
Application example:
• Nitrogen has three half filled pure orbitals
    ♦ So it can form NH₃
    ♦ But it prefers to use hybrid orbitals
    ♦ See section 4.28
4. The hybrid orbitals push each other as far away as possible
    ♦ This is to reduce repulsion
    ♦ So they orient in some specific directions
    ♦ Since there are specific directions, we are able to find the shapes of various molecules
Application example:
• As a result of hybridization in CH₄, four hybrid orbitals are produced
    ♦ Each of those hybrid orbitals try to push the other as far away as possible
    ♦ This ‘pushing’ results in a tetrahedral shape
    ♦ Thus the shape of CH₄ is tetrahedral

---

Now we will write the four important conditions for hybridization

1. The orbitals present in the valence shell of the atom are hybridized
Application:
• We need not consider the inner orbitals while analyzing hybridization in an atom
• We need to consider only the valence shell orbitals
2. The orbitals undergoing hybridization should have almost equal energy
Application example:
• In PCl₅, we saw that 4s, 3p and 3d of P cannot hybridize together
    ♦ This is because, their energies differ vastly
    ♦ See fig.4.176 of the previous section
3. Promotion of electron is not an essential condition prior to hybridization
Application example:
• In PCl₅, we saw that, one electron in the 3s orbital gets promoted to a 3d orbital    
    ♦ This promotion may take place even after the formation of the five hybrid orbitals
4. It is not necessary that, only half filled orbitals participate in hybridization. In some cases, even filled orbitals in the valence shell take part in hybridization
Application example:
• In NH₃, the 2s orbital is completely filled
    ♦ Even then it participates in hybridization
    ♦ See fig.4.160 in section 4.28

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Now we can write the definition of hybridization. It can be written in 2 steps
1. Hybridization is the process of intermixing of orbitals
    ♦ Those participating orbitals should have only ‘slightly different’ energies
    ♦ During the hybridization, ‘redistribution of energies’ take place
2. After hybridization, a set of new orbitals is produced
    ♦ Each of those new orbitals will have the same energy
    ♦ Each of those new orbitals will have the same shape

---

• We have completed a discussion on valence bond theory and hybridization concept
• We will see the solved examples (section 4.40) after completing a discussion on molecular orbital theory also
• In the next few sections, we will be discussing the molecular orbital theory

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Chapter 4.30 - Structure of Phosphorus Pentachloride

• In the previous section 4.29, we saw the structure of BCl₃ and BeCl₂. In this section, we will see the hybridization involving d orbitals

• In the cases that we saw so far, the hybridization of the central atom involved s and p orbitals only
• When we consider the elements in the third period, we will have to deal with d orbitals also
    ♦ That means, the d orbitals will also mix together with s and p orbitals
■ For example:
• In sp³d hybridization, there will be  one s, three p and one d orbitals
    ♦ That is., (s + p + p + p + d)
• Let us see the various possibilities when d orbitals are also involved in the hybridization. It can be written in 3 steps:
1. Fig.4.173 below shows the energy levels of various orbitals
(Note: The actual names of the five d-orbitals are: $\mathbf\small{d_{xy},\;d_{xz},\;d_{yz},\;d_{x^2-y^2},\;\rm{and}\;d_{z^2}}$. In the figs. below, they are named as d_i, d_ii, d_iii, d_iv and d_v. This is for saving space)  

Fig.4.173

2. Let us compare some energies
• The lowest d-orbitals are the 3d orbitals. So we will investigate the comparisons related to those 3d orbitals:
• We see that:
(i) The energy of 3d is comparable to the energy of 3s and 3p
(ii) The energy of 3d is again comparable to the energy of 4s and 4p
(iii) The energy of 3d is not comparable to the energy of 4s and 3p
    ♦ This is because, there is much difference between the energies of those orbitals
          ✰ The 4s sub-shell is in a 'higher main-shell'
          ✰ The 3p and 3d sub-shells are in a 'lower main-shell'  
3. Based on the above comparison, we can write the 'combinations which are possible' and 'combinations which are not possible'
(i) The following two combinations are possible:
    ♦ Hybridization involving 3s, 3p and 3d orbitals [based on 2(i)]
          ✰ This is shown in fig.4.174 below
    ♦ Hybridization involving 3d, 4s and 4p orbitals [based on 2(ii)]
          ✰ This is shown in fig.4.175 further below
(ii) The following combination is not possible:
    ♦ Hybridization involving 3p, 3d and 4s [based on 2(iii)]
          ✰ This is shown in fig.4.176 further below
          ✰ The outline is drawn using dashed lines to indicate that, the combination is not possible

Fig.4.174

Fig.4.175

Fig.4.176
This combination is not possible

---

Let us see some real molecules in which, the d orbitals also participates in the hybridization

Structure of PCl₅
• When we discussed VSEPR theory, we saw the Lewis dot structure of PCl₅
    ♦ See fig.4.93(c) in section 4.15
    ♦ All the five P-Cl bonds are single bonds
• For convenience, that fig. is shown again below:

• The details about the shape of PCl₅ can be written in 12 steps:
1. The P atom in PCl₅ is sp³d hybridized
• Let us see how this sp³d hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.177(a) below shows the orbitals of P

Fig.4.177

(ii) When enough energy is given, one electron in the 3s orbital jumps to the 3d_i orbital
• Thus we get five half filled orbitals. This is shown in fig.b
(iii) The 3s orbital mixes together with 3p_x, 3p_y, 3p_z and 3d_i. This is also shown in fig.b
(iv) Since there is one s-orbital, three p-orbitals, and one d orbital, it is a sp³d hybridization
2. We know that:
    ♦ In sp³d hybridization, there will be five resulting hybrid orbitals
• Together, they form a triangular bipyramidal shape
    ♦ This can be explained in 3 steps:
(i) Fig.4.178(a) below, shows the outlines of a triangular bipyramid

Fug.4.178

    ♦ The base triangle is shown in blue dashed lines
          ✰ The small red sphere is the nucleus of the P atom
          ✰ This nucleus is situated a the 'center of gravity' of the base triangle
    ♦ Three green dashed lines radiate out from the nucleus
          ✰ These lines are at an angular distance of 120^o apart
    ♦ Two red dashed lines radiate upwards and downwards from the nucleus
          ✰ These lines together form the axis of the bipyramid
    ♦ The magenta dashed lines are the sloping edges of the bipyramid
(ii) Lines which are required:
    ♦ We require the following lines:
          ✰ The three green dashed lines
          ✰ The two red dashed lines (these two, together form the axis)
    ♦ We do not require the following lines:
          ✰ The six magenta dashed lines
          ✰ The three blue dashed lines
(iii) The required lines are shown in fig.4.178(b)
    ♦ Altogether, there are five lines
    ♦ The five hybrid orbitals will be oriented along these lines
    ♦ This is shown in fig.c
■ Thus we get the 'sp³d hybridized P atom'
3. Let us write three 'important points to remember' about fig.4.178:
(i) The 'plane of the base triangle' is exactly at the middle of the structure
    ♦ So this plane is called 'equatorial plane'
(ii) All the three green dashed lines lie on the 'plane of the base triangle'
    ♦ So the three orbitals along the green lines are called 'equatorial orbitals' 
    ♦ These orbitals are at an angular distance of 120^o apart
(iii) The two red dashed lines form the axis of the structure
    ♦ So the two orbitals along the red lines are called 'axial orbitals' 
    ♦ They are at an angular distance of 180^o apart
4. So we can write:
    ♦ The 'sp³d hybridized P atom' consists of two items:
          ✰ The five hybrid orbitals
          ✰ The nucleus (small red sphere)
 • In the fig.4.178(c), the 'smaller lobes of hybrid orbitals' are not shown
    ♦ This is to reduce congestion and thus obtain greater clarity
5. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp³d hybrid orbitals are formed from 'one s orbital', 'three p orbitals' and 'one d orbital'
(ii) In our present case of P atom, they are: 'one 3s orbital', 'three 3p orbitals' and 'one 3d orbital'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of five electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the three electrons?
Answer:
• The five electrons will be distributed among the five sp³d hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.178(c) above

6. So in fig.4.178(c), we have:

    ♦ Five half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. It can be written in 2 steps:
(We have already seen the two steps in the previous section. But we will write them again)
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron
• We saw it in fig.4.166 in the previous section. It is shown again below:

Fig.4.166

• The p_x orbital is shown in red color
    ♦ It has two arrows
• The p_y orbital is shown in green color
    ♦ It has two arrows
• The p_z orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
7. One Cl atom will come and overlap with each of the five half filled orbitals of the P atom
• So a total of five Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.179(a) below:

Fig.4.179

• In the fig.4.179(a) above, the 'p_z orbitals of Cl' have become completely filled
8. The electron clouds of Cl:
(i) The red, green, blue and cyan regions around the Cl nucleus are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(ii) So we do not need to show the orbitals (electron clouds) of the Cl atoms. We can delete them
• However, we will retain the pz orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the P atom
(iii) The final structure is shown in fig.4.179(b) above
• The small yellow spheres represent the nuclei of Cl atoms

9. Thus we get the model of the PCl₅ molecule
• Let us write the salient features of this model. It can be written in 3 steps:
(i) There is a total of five P-Cl bonds
    ♦ Three of them lie on the equatorial plane
    ♦ The remaining two lie along the 'axis of the bipyramid'
(ii) So there are:
    ♦ Three equatorial bonds
    ♦ Two axial bonds
(iii) The angle between
    ♦ Any two equatorial bonds is 120^o
    ♦ The two axial bonds is 180^o
    ♦ Any equatorial bond and axial bond is 90^o
10. Now we will see an interesting feature about bond lengths
(i) Consider fig.4.178(b) that we saw earlier
    ♦ There is a total of 5 dashed lines
    ♦ Those five lines give us the directions of the bonds
(ii) When the PCl₅ molecule is formed, each of those five lines will contain a bond pair of electrons
    ♦ Repulsion will occur between those electron pairs
    ♦ We can mark those repulsions in the fig.4.178(b)
    ♦ Fig.4.180(a) below, is one such fig.
          ✰ It shows the repulsion in the equatorial plane (horizontal plane)
    ♦ Fig.4.180(b) is another such fig.
          ✰ It shows the repulsion in the vertical planes

Fig.4.180

(iii) We can ignore fig.a
• Because, the red double headed arrows will cancel each other out
(iv) But we cannot ignore fig.b
• Let us write the reason. It can be written in three steps (a), (b) and (c):
    (a) There are three blue arrows above the equatorial plane
          ✰ Those arrows will push the 'top most electron pair' away from the equatorial plane
          ✰ So that electron pair will move further upwards
          ✰ As a result, the 'bond length' of the 'upper axial bond' will increase
    (b) There are three blue arrows below the equatorial plane
          ✰ Those arrows will push the 'bottom most electron pair' away from the equatorial plane
          ✰ So that electron pair will move further downwards
          ✰ As a result, the 'bond length' of the 'lower axial bond' will increase
(c) The difference in bond lengths is shown in fig.4.180(c) above 
11. Fig.4.180(c) shows the 2D representation also
• Recall that, we saw the same 2D representation when we analysed PCl₅ using VSEPR theory. See fig.4.93(c)  in section 4.15
• There we saw the reason for giving the solid triangle, dashed triangle etc.,
12. Note that, all the bonds in PCl₅ are sigma bonds

Structure of PF₅

1. We saw the trigonal bipyramidal structure of PCl₅
    ♦ PF₅ also has the same trigonal bipyramidal structure
    ♦ That means, PF₅ will have the same structure as shown in fig.4.179(b) above
2. Note that Cl and F belongs to the same group in the periodic table
    ♦ The electronic configuration of Cl is [Ne]3s² 3p_x² 3p_y² 3p_z¹
    ♦ The electronic configuration of F is [He]2s² 2p_x² 2p_y² 2p_z¹
3. Now the reader can write the necessary steps to arrive at the structure of PF₅
[Hint: In the case of PF₅, the blue orbitals in fig.4.179(b) will be 2p_z of F atoms]

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• In the next section, we will see the structure of SF₆ (Sulfur hexaflouride)

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Chapter 4.29 - Structure of Trichloro boron and Beryllium chloride

• In the previous section 4.28, we saw the structure of NH₃ and H₂O. In this section, we will see the structure of BCl₃ (Trichloro boron) and BeCl₂ (Beryllium chloride)

Structure of BCl₃
• Fig.4.164(a) below shows the Lewis dot structure of BCl₃
• The details about the model of BCl₃ can be written in 10 steps:
1. The B atom in BCl₃ is sp² hybridized
• Let us see how this sp² hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.164(b) below shows the orbitals in the valence shell of B

Fig.4.164

(ii) When enough energy is given, one electron in the 2s orbital jumps to the 2p_y orbital
• Thus we get three half filled orbitals. This is shown in fig.c
(iii) The 2s orbital mixes together with 2p_x and 2p_y. This is shown in fig.d
(iv) Since there is one s-orbital and two p-orbitals, it is a sp² hybridization
2. We know that:
    ♦ In sp² hybridization, there will be three resulting hybrid orbitals
          ✰ Together, they form a triangular planar shape
          ✰ The larger lobes will be directed towards the corners of a triangle
• This is shown in fig.4.165 below:

Fig.4.165

• Let us write four 'important points to remember' about fig.4.165:
(i) The three sp² hybrid orbitals are at an angular distance of 120^o apart 
(ii) The nucleus of the B atom is shown as a small red sphere
    ♦ This sphere is situated at the origin of the coordinate axes
(iii) One of the orbitals lies exactly along the x-axis (red axis)
(iv) We know that, all the three sp² hybrid orbitals will lie on a plane
    ♦ So in fig.4.165, all the three orbitals are lying on the xy-plane
3. So we can write:
    ♦ The 'sp² hybridized B atom' consists of two items:
          ✰ The three hybrid orbitals
          ✰ The nucleus (small red sphere)
 • In the fig.4.165, the 'smaller lobes of hybrid orbitals' are not shown
    ♦ This is to reduce congestion and thus obtain greater clarity
4. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp² hybrid orbitals are formed from 'one s orbital' and 'two p orbitals'
(ii) In our present case of B atom, they are: 'one 2s orbital' and 'two 2p orbitals'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of three electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the three electrons?
Answer:
• The three electrons will be distributed among the three sp² hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.165 above

5. So in fig.4.165, we have:

    ♦ Three half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. it can be written in 2 steps:
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron
• This is shown in fig.4.166 below:

Fig.4.166

• The p_x orbital is shown in red color
    ♦ It has two arrows
• The p_y orbital is shown in green color
    ♦ It has two arrows
• The p_z orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
6. One Cl atom will come and overlap with each of the three half filled orbitals of the B atom
• So a total of three Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.167 below:

Fig.4.167

• In the fig.4.167 above, the 'p_z orbitals of Cl' have become completely filled
7. Is the structure 3D or 2D?
The answer can be written in 5 steps:
(i) The structure in the above fig.4.167, appears to be a 3D structure
• Because, the red orbitals of the Cl atoms are protruding above and below the xy-plane
(ii) But the red orbitals are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(iii) We see that, all the three Cl nuclei lie in the same plane as the nucleus of B
(Remember that, the nucleus of a Cl atom is at the center of the cyan sphere)
(iv) So the structure is planar. In other words, it is a 2D structure
• We do not need to show the orbitals (electron clouds) of the Cl atoms. So we can delete them
• However, we will retain the p_z orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the B atom
(v) The final structure is shown in fig.4.168(a) below.
• The small yellow spheres represent the nuclei of Cl atoms

Fig.4.168

8. Thus we get the model of the BCl₃ molecule
• Let us write the salient features of this model. It can be written in 3 steps:
(i) The three Cl atoms are situated at the three corners of a triangle
(ii) The B atom is situated at the 'center of gravity' of the triangle
(iii) The angle between any two bonds is 120^o
9. The 2D representation of the BCl₃ molecule is shown on fig.4.168(b) above
10. Note that, all the bonds in BCl₃ are sigma bonds

Structure of BeCl₂
• Fig.4.169(a) below shows the Lewis dot structure of BeCl₂
• The details about the model of BeCl₂ can be written in 10 steps:
1. The Be atom in BeCl₂ is sp hybridized
• Let us see how this sp hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.169(b) below shows the orbitals in the valence shell of Be

Fig.4.169

(ii) When enough energy is given, one electron in the 2s orbital jumps to the 2p_x orbital
• Thus we get two half filled orbitals. This is shown in fig.c
(iii) The 2s orbital mixes together with 2p_x. This is shown in fig.d
(iv) Since there is one s-orbital and one p-orbital, it is a sp hybridization
2. We know that:
    ♦ In sp hybridization, there will be two resulting hybrid orbitals
          ✰ Together, they form a linear shape
• This is shown in fig.4.170 below:

Fig.4.170

• Let us write four 'important points to remember' about fig.4.170:
(i) The two sp hybrid orbitals are at an angular distance of 180^o apart 
(ii) The nucleus of the Be atom is shown as a small red sphere
    ♦ This sphere is situated at the origin of the coordinate axes
(iii) Both the orbitals lie exactly along the x-axis (red axis)
(iv) We know that, each hybrid orbital have a larger lobe and a smaller lobe
• In the previous cases (except C₂H₂), we deliberately chose not to show the smaller lobes. This was for better clarity
• In our present case, we do not have to hide the smaller lobes because, they are automatically hidden
    ♦ The smaller lobe of the left side orbital is inside the larger lobe of the right side orbital
    ♦ The smaller lobe of the right side orbital is inside the larger lobe of the left side orbital
(Recall that, in C₂H₂ also, we encountered the same situation)
3. So we can write:
    ♦ The 'sp hybridized B atom' consists of two items:
          ✰ The two hybrid orbitals
          ✰ The nucleus (small red sphere)
4. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp hybrid orbitals are formed from 'one s orbital' and 'one p orbital'
(ii) In our present case of Be atom, they are: 'one 2s orbital' and 'one 2p orbital'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of two electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the two electrons?
Answer:
• The two electrons will be distributed among the two sp hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.170 above

5. So in fig.4.170, we have:

    ♦ Two half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. it can be written in 2 steps:
(We have already seen the two steps when we discussed BCl₃ above. But we will write them again)
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron. This was shown in fig.4.166 above

• The px orbital is shown in red color
    ♦ It has two arrows
• The py orbital is shown in green color
    ♦ It has two arrows
• The pz orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
6. One Cl atom will come and overlap with each of the two half filled orbitals of the Be atom
• So a total of two Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.171 below:

Fig.4.171

• In the fig.4.171 above, the 'p_z orbitals of Cl' have become completely filled
7. Is the structure 3D or 2D?
The answer can be written in 5 steps:
(i) The structure in the above fig.4.171, appears to be a 3D structure
• Because, the red orbitals of the Cl atoms are protruding above and below the xy-plane
(ii) But the red orbitals are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(iii) We see that, all the three Cl nuclei lie in the same plane as the nucleus of B
(Remember that, the nucleus of a Cl atom is at the center of the cyan sphere)
(iv) So the structure is planar. In other words, it is a 2D structure
• We do not need to show the orbitals (electron clouds) of the Cl atoms. So we can delete them
• However, we will retain the p_z orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the Be atom
(v) The final structure is shown in fig.4.172(a) below.
• The small yellow spheres represent the nuclei of Cl atoms

Fig.4.172

8. Thus we get the model of the BeCl₂ molecule
• Let us write the salient features of this model. It can be written in 4 steps:
(i) The two Cl atoms are situated at the opposite ends of a line
(ii) The Be atom is situated at the 'center of gravity' (midpoint) of the line
(iii) The angle between the two bonds is 180^o
(iv) So it is a linear structure
9. The 2D representation of the BeCl₂ molecule is shown on fig.4.172(b) above
10. Note that, both the bonds in BeCl₂ are sigma bonds

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• In the next section, we will see the hybridization involving d orbitals

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